Self-sulfate sulfate and other ions

[Weissritter]

First disclaimer: I do not want this for homework. I simply want an answer.
Nao, what I've seen, a single halogen can get into a 0 oxidation state combining with itself.
Now...could an inorganic ion have a total 0 by combining with itself? Could a SO4 join with another and create an (SO4)-(SO4), like a self sulfate?
For now, I have heavy evidence suspecting a big, huge 'NO!' and comments sending me to pay attention to my Chem class.
But, "Be brave, the bravest ever.". So, what do our Chem pros think about a self sulfate or self nitrate and relative compounds?

 

[AGNuke]

Marshall's acid - S₂O₈^2-.

It sulphate bonded to another sulphate doesn't mean that its Oxidation number is going to be zero.

And what you interpreted for halogens is incorrect. A Halide ion, X^- is way different that Halogen, X₂. Halide ions DO NOT combine with themselves to form zero ON Halogen. First they lose their excess electron to become a single halogen atom, Then they combine with other halogen atom to form a stable halogen molecule.

And really, pay some attention. You interpreted your initial statement wrong. You need to consider the electron.$$2X^- \rightarrow X_2 + 2 e^-$$
See, electrons are ejected from the overall reaction. So, Oxidation number is not becoming zero just because they are combining with themselves, but because they are losing their excess electron and getting themselves to zero ON before combining.

 

[JohnRC]

AGNuke's reply is right on. However, something that might interest the OP is that there is a group of anions, sometimes called "pseudo-halogens", that can undergo a 2-electron oxidation to produce a "oxidation state 0" dimer.

The most obvious example is the hydroxyl ion OH^–:

2 OH^– → HOOH + 2 e^–

Cyanide ion/cyanogen is another example

2 CN^– → NCCN+ 2 e^–

Note that in cases like these neither of the elements goes to oxidation state zero, only the radical as a whole. In hydrogen peroxide hydrogen is in oxidation state +1 and oxygen in oxidation state –1; it is only by looking at an ^•OH radical as a whole that we can arrive at oxidation state 0.

Note also that neither sulfate nor nitrate is among these pseudohalogen anions

 

[Weissritter]

And what you interpreted for halogens is incorrect. A Halide ion, X^- is way different that Halogen, X₂. Halide ions DO NOT combine with themselves to form zero ON Halogen. First they lose their excess electron to become a single halogen atom, Then they combine with other halogen atom to form a stable halogen molecule.

And really, pay some attention. You interpreted your initial statement wrong. You need to consider the electron.$$2X^- \rightarrow X_2 + 2 e^-$$
See, electrons are ejected from the overall reaction. So, Oxidation number is not becoming zero just because they are combining with themselves, but because they are losing their excess electron and getting themselves to zero ON before combining.

Wait.
What the actual phenomenon happens when a carbon in methane becomes a carbon in CO₂? Carbon has a starting -4 in methane, then "loses" EIGHT electrons when becoming +4 in CO₂. An usual carbon has six electrons and six protons.
Just how does this happen?

AGNuke's reply is right on. However, something that might interest the OP is that there is a group of anions, sometimes called "pseudo-halogens", that can undergo a 2-electron oxidation to produce a "oxidation state 0" dimer.

The most obvious example is the hydroxyl ion OH^–:

2 OH^– → HOOH + 2 e^–

Cyanide ion/cyanogen is another example

2 CN^– → NCCN+ 2 e^–

Note that in cases like these neither of the elements goes to oxidation state zero, only the radical as a whole. In hydrogen peroxide hydrogen is in oxidation state +1 and oxygen in oxidation state –1; it is only by looking at an ^•OH radical as a whole that we can arrive at oxidation state 0.

Note also that neither sulfate nor nitrate is among these pseudohalogen anions

And yes, I had no idea these group of pseudo-halogens existed. Luckily I found a short list with them. But it has only four. Do you have a more complete one?

 

[JohnRC]

Nature knows nothing about "oxidation numbers". They are part of a chemist's artificial accounting scheme.

If you burn methane, here is (part of) what happens on a molecular scale:

^•OH + CH₄ → H₂O + ^•CH₃ (C from –4 to –3)

^•CH₃ + O₂ → CH₃OO^• (C from –3 to –2)

CH₃OO^• → CH₂O + ^•OH (C from -2 to 0)

^•OH + CH₂O → ^•CHO + H₂O (C from 0 to +1)

^•CHO + O₂ → CO + HO₂^• (C from +1 to +2)

CO + HO₂^• → CO₂ + ^•OH (C from +2 to +4)

Combustion reactions with oxygen are extremely complicated; any serious discussion of methane oxidation would involve at least about 50 steps like this. I have just chosen a straightforward path through to the product with reasonably plausible steps.

The important thing is that the reaction steps are progressive, and that you can trace the transfer of 8 electrons in going from reactants to products using the chemist's accounting scheme.

 

[JohnRC]

Moeller, "Inorganic Chemistry" (1952-1959) devotes a large section of his textbook to what he calls "halogenoids" (pp. 462-480). He mentions cyanide ion/cyanogen, thiocyanate ion/thiocyanogen, selenium and tellurium analogues of thiocyanate, and thiocarbamate ion/azidocarbondisulfide. He does not claim that his list is exclusive. He also suggests that cyanate ion and azide ion might belong to this family, but (OCN)₂ is unknown, while N₆ is simply bizarre!